abs(A+B)^2 = ?

[Guest]

Hello.

What is abs(A+B)^2 equal to?
where we have:
A = R exp(i*a)
and
B = S exp(i*b)

I would like a small and simple proof too if possible.

Thanks in advance for your answers.

[Guest]

The answer is [tex]R^2 + S^2+2R S \cos(a-b)[/tex].

You can get this answer by drawing the Argand diagram, you should get something similar to
http://image.tutorvista.com/cms/images/ ... graph.jpeg
Here [tex]|OP| = R, |PR| = S, |OR| = |A+B|, \angle OPR = \pi +a-b[/tex] (sorry the labelling is a bit confusing as the diagram uses the letter R as well, but hopefully it should be clear from the context when I mean [tex]R[/tex] from the diagram and [tex]R[/tex] from your question). Then to get the answer it is a just a simple application of the cosine rule from trigonometry (and the fact that [tex]\cos(\pi+\theta) = -\cos\theta[/tex], where [tex]\theta = a-b[/tex] in our case).

Another simple way to get the answer is to just expand out [tex]A[/tex] and [tex]B[/tex] using Euler's formula, and use the trigonometry identities [tex]\cos^2\theta+\sin^2\theta=1[/tex] and [tex]\cos(a-b) = \cos a \cos b +\sin a \sin b[/tex].
[tex]|A+B|^2[/tex]
[tex]= |Re^{ia}+Se^{ib}|^2[/tex]
[tex]= |R\cos a + iR\sin a +S\cos b +i S\sin b|^2[/tex]
[tex]= |(R\cos a+S\cos b)+i(R\sin a+S\sin b)|^2[/tex]
[tex]= (R\cos a+S\cos b)^2+(R\sin a+S\sin b)^2[/tex]
[tex]= R^2\cos^2 a + 2RS\cos a \cos b + S^2 \cos^2 b + R^2\sin^2 a+2RS\sin a\sin b +S^2\sin^2 b[/tex]
[tex]= R^2(\cos^2 a+\sin^2 a) + S^2(\cos^2 b +\sin^2 b) + 2RS(\cos a \cos b+\sin a\sin b)[/tex]
[tex]= R^2+S^2+2RS\cos(a-b)[/tex]

Hope this helped,

R. Baber.

[Guest]

Another way you could calculate it is by using complex conjugates (together with Euler's formula and the fact that [tex]\cos(-\theta) = \cos\theta[/tex], and [tex]\sin(-\theta)=-\sin\theta[/tex]).

[tex]|A+B|^2[/tex]
[tex]= (A+B)\overline{(A+B)}[/tex]
[tex]= (A+B)(\overline{A}+\overline{B})[/tex]
[tex]= (Re^{ia}+Se^{ib})(Re^{-ia}+Se^{-ib})[/tex]
[tex]= R^2e^{i(a-a)}+S^2e^{i(b-b)}+RS(e^{i(a-b)}+e^{i(b-a)})[/tex]
[tex]= R^2+S^2+RS(e^{i(a-b)}+e^{-i(a-b)})[/tex]
[tex]= R^2+S^2+RS(\cos(a-b)+i\sin(a-b)+\cos(-(a-b))+i\sin(-(a-b)))[/tex]
[tex]= R^2+S^2+RS(\cos(a-b)+i\sin(a-b)+\cos(a-b)-i\sin(a-b))[/tex]
[tex]= R^2+S^2+2RS\cos(a-b)[/tex]

R. Baber.